Let x = y = 0 on plane 6z = 2 - 3x + 9
6z = 2
z = 3
Thus we have the point (0,0,3) and to find the distance between this point on Plane B and Plane A there is the formula:

absolute value( a(x0) + b(y0) + c(z0) - D ) / sqrt(a^2 + b^2 + c^2)

plane A has: a = 2, y = -6, z = 4, d = 0

so, i did 2(0) - 6(0) + 4(3) = 12 / sqrt( 4 + 36 + 16)

12 / sqrt(56) and further reduced to 6/sqrt(14)

the computer assignment says this is wrong... any help is greatly appreciated.

I agree with your answer. But try this form:

May 27th 2010, 05:11 PM

sj9110

What's the difference? (how did you arrive at that answer?).. also, it was a test so I only had that one shot, what I really want to know is if I have a legitimate claim that I deserve points for this problem so I can bring it up with my professor..

May 27th 2010, 09:05 PM

11rdc11

Quote:

Originally Posted by sj9110

What's the difference? (how did you arrive at that answer?).. also, it was a test so I only had that one shot, what I really want to know is if I have a legitimate claim that I deserve points for this problem so I can bring it up with my professor..

rationalize the den.

My teacher marks the answer wrong unless it is the simplest form, so I do not think he will give you credit.

May 28th 2010, 03:30 AM

HallsofIvy

Quote:

Originally Posted by sj9110

I had the following problem on my test today and I supposedly got the wrong answer but I can't seem to find where I went wrong. Please let me know if you see an error in my work:

Thus we have the point (0,0,3) and to find the distance between this point on Plane B and Plane A there is the formula:

absolute value( a(x0) + b(y0) + c(z0) - D ) / sqrt(a^2 + b^2 + c^2)

plane A has: a = 2, y = -6, z = 4, d = 0

so, i did 2(0) - 6(0) + 4(3) = 12 / sqrt( 4 + 36 + 16)

12 / sqrt(56) and further reduced to 6/sqrt(14)

the computer assignment says this is wrong... any help is greatly appreciated.

Thanks in advance.

You really shouldn't need to memorize a formula for this- pick any point on the first plane- (0, 0, 0) is an obvious choice. Then construct the line through that point perpendicular to the two planes. The two planes can be written x- 3y+ 2z= 0 and x- 3y+ 2z= 2/3 so a normal line through (0, 0, 0) is x= t, y= -3t, z= 2t. That intersects the second plane when t- 3(-3t)+ 2(2t)= (1+ 9+ 4)t= 14t= 2/3 or t= 1/21. The point of intersection is (1/21, -1/7, 2/21).

The distance between the planes is the distance from (0, 0, 0) to (1/21, -1/7, 2/21).